Let the internal angle bisector of $A$ intersect the circumcircle of $ABC$ again at $X$, and let the external angle bisector intersect the circumcircle of $ABC$ again at $X'$. I claim that $X'$ lies on the circumcircle of $ADM$. Proof: It is well known that $X'$ lies on the perpendicular bisector of $BC$, and that $AX'$ is perpendicular to $AD$. It follows that $\angle DAX' = \angle X'MD = 90$ and hence $ADMX'$ is cyclic. Not only that, but the circumcentre of $ADMX'$ is the midpoint of $X'D$, and the circumcentre of $ABC$ is the midpoint of $X'X$. Then by the midpoint theorem, we are done.

This problem was proposed by Burii.

Let the circumcircle of $ADM$ intersects the circumcircle of $ABC$ at $E$ other than $A$. It is well known that as $AE$ is the radical axis of both circles, then the line through centers of circumcircles of $ABC$ and $ADM$ is perpendicular to $AE$. It is enough to prove $AD$ is perpendicular to $AE$. Now we use phantom point. Let $E'$ be the point where the perpendicular bisector of $BC$ meet the circumcircle of $ABC$ that is in the middle of arc $BC$ containing $A$. Then because $E'M$ pass through the center of the circumcircle of $ABC$, and $AD$ meet the circumcircle of $ABC$ at the midpoint of arc $BC$ not containing $A$, then $\angle EAD=90$. We also know that $\angle E'MD=90$, so we get $AE'MD$, is a cyclic quadrilateral. Because $E'$ is a point on the circumcircle of $ABC$, $E'=E$. So, we get $\angle DAE=90$ or $AD \perp AE$. Because of the line through the centers of circumcircles of $ABC$ and $ADM$ is perpendicular to $AE$, and $AD \perp AE$, obviously $AD \parallel$ line through centers of circumcircles of $ABC$ and $ADM$, and we're done.

Very easy but nice Let $N$ be the midpoint of arc $BAC$. Clearly $\angle NAD = \angle NMD = 90^{\circ}\implies N\in\odot(ADE)$. Hence $AN$ is the common chord of $\odot(ABC), \odot(ADE)$, which is perpendicular to $AD$ so we are done.

MarkBcc168 wrote: Very easy but nice Let $N$ be the midpoint of arc $BAC$. Clearly $\angle NAD = \angle NMD = 90^{\circ}\implies N\in\odot(ADE)$. Hence $AN$ is the common chord of $\odot(ABC), \odot(ADE)$, which is perpendicular to $AD$ so we are done. Why is it clear that $\angle NAD = \angle NMD$?

It's clear that $MN\perp BC$ so $\angle NMD = 90^{\circ}$. Moreover, $AN$ is external angle bisector of $\angle BAC$ so $\angle NAD = 90^{\circ}$.

Let $L$ be the midpoint of arc $BAC$, the circumcenter of $ABC$ be $O_1$, and the circumcenter of $ADM$ be $O_2$. Thales' implies $ADML$ is cyclic with diameter $DL$. Hence, properties of Radical Axes give $$AD \perp AL \perp O_1O_2$$which suffices. $\blacksquare$

Let $E=AD\cap (ABC)$ and $X=OE\cap (ADM)$ and $O'=$ center of $(ADM)$. Since $ADMX$ is cyclic $\angle DAX=90$ $\implies$ $X\in (ABC)$ $\implies$ $AX$ is radical axis of $(ABC)$ and $(ADM)$. So $OO'\perp AX\perp AD$ $\implies$ $AD||OO'$.

Let K be midpoint of arc BAC and O1 be circumcenter of ABC and O2 be circumcenter of ADM. we know ADMN and ABCN both are cyclic and AN is their common chord so O1O2 is perpendicular to AN. we know AD is also perpendicular to AN so O1O2 || AD and we're Done.

Let $N$ be the midpoint of arc $\widehat{BC}$ containing $A.$ Notice $\angle DAN=90=\angle NMD$ so $ADMN$ is cyclic. Hence, $\ell\perp\overline{AN}\perp\overline{AD}$ where $\ell$ is the line connecting the centers. $\square$

Let $O$ be the circumcenter of triangle $ABC$, $F$ be the circumcenter of triangle $ADM$. Let $AD$ intersect $OM$ at $E$. Let the two circumcircles intersect again at $G$. Note that if the 2 circles only intersect once at $A$, then $A,F,O,D$ are collinear and we are done. $\angle EGA=\angle ECA=180^{\circ} -\angle CEA- \angle CAE=180^{\circ} -\angle CBA-\angle DAB= \angle ADB= \angle CDE=\angle MGA$, so $E,M,G$ collinear. Since $E,M,O$ collinear, $E,M,O,G$ collinear. Then $\angle EAG=90^{\circ}$. But note that $OF$ is the radical axis of the two circles, so both $OF$ and $EA$ are perpendicular to $AG$. Hence, we have $OF\parallel EA$.